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anonymous
 4 years ago
Where is the gradient of phi
a) perpendicular
b) parallel
to the zaxis?
anonymous
 4 years ago
Where is the gradient of phi a) perpendicular b) parallel to the zaxis?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I guess I have to set something zero for a) and perhaps in b) z should be constant?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0use dot product. If need more help tell. :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I do the dot thing to get the parellel part and a cross product for the perpendicular part?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes. to find where it is perpendicular to z axis solve this equation: nabla f . (0,0,1)=0 for paralel : nabla f . (0,0,1)= nabla f

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0other way would be finding maximum and minimum of nabla f

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So for the parallel part I dont use a any vector but the vector (0,0,1) and dot this vector with nabla?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for a) I ended up with z^2=xy Is that my final answert for a set of points for which the gradient is parallel to the zaxis?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for b) I got now y^2xz=x^2yz This looks also not very good as a final answer.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for b): 3(x^2yz,y^2xz,z^2xy).(0,0,1)=3sqrt((x^2yz)^2+(y^2xz)^2+(z^2xy)^2)=> (z^2xy)=sqrt((x^2yz)^2+(y^2xz)^2+(z^2xy)^2) now should try to simplify this somehow

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0noticing that it will happen when : (x^2yz)^2=0 and (y^2xz)^2=0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so x^2=yz and y^2=xz from 1º: y=x^2/z putting this in 2º: x^4/z^2=xz => x^3=z^3 =>x=z=y which is a line passing through the origin forming 45º with every axis

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So x, y, and z can be any number but as long as the have all the same value the gradient of phi is parallel to the zaxis?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0:( c) When is the gradient zero. There I already calculated that the gradient would be zero for x=y=z so it seems kind of weird that this is the same solution for the gradient to be parallel to the zaxis.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But you calculated x=y=z so how can x=y=0 and z= anything?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ya that's true. Forget it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0On the other hand as much as I remmeber vector that is 0 is also considered perpendicular or parallel to anything

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This is confusing. Did I at least got a) right? I ended up with z^2=xy but that looks kind of weird for a final answert, but I don´t know if I can simplify or clarify that anymore.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes it's right. Here you got the graph of that: http://www.wolframalpha.com/input/?i=z%5E2%3Dxy+

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think part b) is ok too. Not sure tough, how to explain that it is same as for c)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hmmmm ... I am still pretty confused but if you think this is alright than I will stick to it. Thank you for your help, especially since this took so long. Thanks a lot, myko.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Glad to help. Sry for not clarifying it till the end...
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